Question

An Organic compound on analysis given the following
Carbon
20%
hydrogen = 6.66%
Oxygen - 53.34%
calculate its molecular formula if its molecular mass is 90​

Answer 1

Correct Question :-

An organic compound on analysis gave the following composition.  Carbon = 40%, Hydrogen = 6.66%, Oxygen = 53.34%. Calculate its molecular formula, if molecular mass is 90.

Solution :-

Given :-

Percentage composition of carbon (C) = 40%

Percentage composition of hydrogen (H) = 6.66%

Percentage composition of oxygen (O) = 53.34%

We know :-

Atomic mass of carbon (C) = 12g

Atomic mass of hydrogen (H) = 1g

Atomic mass of oxygen (O) = 12g

$\bullet \ \sf No \ of \ moles \ of \ carbon \ (C) = \dfrac{40}{12} = 3.33$

$\sf\bullet \ No \ of \ moles \ of \ hydrogen \ (H) = \dfrac{6.66}{1}= 6.66$

$\bullet \sf \ No \ of \ moles \ of \ oxygen \ (O) = \dfrac{53.34}{16}= 3.33$

Simplest ratio :-

$\longrightarrow \sf \dfrac{3.33}{3.33} \ : \ \dfrac{6.66}{3.33} \ : \ \dfrac{3.33}{3.33} \\\\\longrightarrow 1 \ : \ 2 \ : \ 1$

→ Empirical formula = $\sf CH_2O$

→ Empirical formula mass = 12 + 2 + 16

                                          = 30g

We know :-

$\bf Molecular \ formula = n \times Empirical \ formula$

Here :-

$\longrightarrow \sf n = \dfrac{Molecular \ mass }{Empirical \ formula \ mass } \\\\\longrightarrow n = \dfrac{90}{30} \\\\\longrightarrow \bf n = 3$

Molecular formula = $\sf 3 \times (CH_2O)$

$\underline{\boxed{\bf Molecular \ formula = C_3H_6O_3}}$